We give a simple brute-force solution for Exercise 4 (a).

(1)

Exercise 4 0 Points

A setais inductive if∅ ∈a and for allx∈a,x∪ {x} ∈a. Letω =^T{x|x is inductive}.

(a) Show that ω is a set.

SolutionThe intersection of every nonempty class is a set. Indeed, let A6=∅be a class, then ^TA = {x | x ∈ y ∈ A for some sety}. As A 6= ∅, there is some set z ∈ A. Then TA={x ∈z |x ∈y∈A for some set y}. Now it is easy to see that every limit stage is inductive, so the class of inductive sets is not empty.

We show: ifs is a limit stage and x ∈sthen x∪ {x} ∈s. As s is a limit stage, we have x ∈s⁰ ∈s for some stage s⁰. Then{x} ⊆s⁰, so x,{x} ∈ P(s⁰). Now we show that if two sets are in some stage then so is their union. The result follows then.

We show first that ifa∈sthen ^Sa∈sfor any set a and any stages. Letb∈^Sa. Then b∈c∈a∈sfor somec∈a. By transitivity ofs,b∈s.

Now,{x,{x}} ∈ P(P(s⁰)) and {x,{x}} ∈s, as sis a limit stage. Then ^S{x,{x}} ∈sand x∪ {x} ⊆^S{x,{x}} ∈s, so{x,{x}} ∈sbecausesis hereditary as a stage.

http://logic.rwth-aachen.de/Teaching/MaLo2-WS10